Spin the octahedron while moving it through a hole

Question

Suppose we have an octahedron with side length a, and we need to make it through a square hole. We can spin the octahedron while moving it, then what is the minimum side length of the square hole?
Edit: "Spin" means we can rotate the octahedron in any ways, choosing random axis.
Edit edit: I just can solve the case when a special axis (the diagonal) is restricted for the rotating... Edit edit edit:I would really appreciate if someone can solve the case that we can vary the rotating axis while moving, I mean, like in real life when we really want to move some thing through a hole, but I guess it would be extremely difficult...

Answer 1

As a lower bound, here is an approach that allows an octahedron of side length $\sqrt{6}-\sqrt{2}\approx 1.03527$ to pass through a unit square:

                                                 

We push the octahedron through face-first, starting with the maximal equilateral triangle that can be inscribed in a square and translating the centroid diagonally as we do so to keep the cross-sections inscribed in the square at all times. The animation shows cross-sections of this process as we push the octahedron through.

Note that this approach only requires translation of the octahedron, and does not make use of any rotation as long as we start in the correct orientation.

I think there is a good chance this construction is optimal, even among strategies that permit arbitrary rotations; one avenue to proving it, if true, would be to show that any cross-section of the octahedron passing through its centroid cannot be fit into a square smaller than the one required for the hexagonal cross-section shown here. (I don't know of a counterexample to this claim, and wouldn't be surprised if it were true, though proving it might still be somewhat tricky.)